5.2 solving systems of linear equations by substitution
learning target: understand how to solve systems of linear equations by substitution
solve the system by substitution. check your solution.
10. ( y = -3x - 7 )
( y = x + 9 )
11. ( \frac{1}{2}x + y = -4 )
( y = 2x + 16 )
12. ( -x + 5y = 28 )
( x + 3y = 20 )
13. zoo admission costs $6 for children and $9 for adults. on monday, 2200 people visit the zoo and the zoo collects $14,850 in admissions.
a. write a system of linear equations that represents this situation.
b. how many zoo visitors are children? adults?
solve the system. explain your choice of method.
14. ( y = x - 2 )
( y = -2x + 1 )
15. ( 3y + 9 = 3x )
( y = \frac{1}{3}x + 1 )
16. ( -x + 2y = -4 )
( 4y = x )
17. the measure of an acute angle in a right triangle is one - fourth the measure of the other acute angle. write a system of linear equations that represents this situation and use it to find the measures of the acute angles of the triangle.
5.3 solving systems of linear equations by elimination
learning target: understand how to solve systems of linear equations by elimination
solve the system by elimination. check your solution.
18. ( 2x + 5y = 60 )
( 2x - 5y = -20 )
19. ( 4x - 3y = 15 )
( 2x + y = -5 )
20. a gift basket that contains jars of jam and packages of bread mix costs $15. there are 8 items in the basket. jars of jam cost $6 each, and packages of bread mix cost $2 each. write and solve a system of linear equations to find the number of each item in the gift basket.
21. when might it be easier to solve a system by elimination instead of graphing?

Question

Accepted Answer

### 5.2 Solving Systems by Substitution
---
#### Problem 10
# Explanation:
## Step1: Substitute $y=x+9$ into first equation
$x+9 = -3x -7$
## Step2: Isolate $x$ terms
$x+3x = -7 -9$
$4x = -16$
## Step3: Solve for $x$
$x = \frac{-16}{4} = -4$
## Step4: Substitute $x=-4$ into $y=x+9$
$y = -4 + 9 = 5$
# Answer:
$x=-4, y=5$

---
#### Problem 11
# Explanation:
## Step1: Isolate $y$ in first equation
$y = -4 - \frac{1}{2}x$
## Step2: Substitute $y$ into second equation
$-4 - \frac{1}{2}x = 2x + 16$
## Step3: Isolate $x$ terms
$-\frac{1}{2}x -2x = 16 + 4$
$-\frac{5}{2}x = 20$
## Step4: Solve for $x$
$x = 20 	imes (-\frac{2}{5}) = -8$
## Step5: Substitute $x=-8$ into $y=-4-\frac{1}{2}x$
$y = -4 - \frac{1}{2}(-8) = -4 +4 = 0$
# Answer:
$x=-8, y=0$

---
#### Problem 12
# Explanation:
## Step1: Isolate $x$ in first equation
$x = 28 -5y$
## Step2: Substitute $x$ into second equation
$28 -5y +3y = 20$
## Step3: Simplify and solve for $y$
$28 -2y =20$
$-2y = 20-28=-8$
$y = \frac{-8}{-2}=4$
## Step4: Substitute $y=4$ into $x=28-5y$
$x=28-5(4)=28-20=8$
# Answer:
$x=8, y=4$

---
#### Problem 13
# Explanation:
## Part a: Define variables and equations
Let $c$ = number of children, $a$ = number of adults.
Total visitors: $c + a = 2200$
Total revenue: $6c +9a =14850$

## Part b: Solve the system
### Step1: Isolate $c$ from first equation
$c=2200 -a$
### Step2: Substitute $c$ into revenue equation
$6(2200 -a) +9a =14850$
### Step3: Expand and simplify
$13200 -6a +9a =14850$
$3a =14850-13200=1650$
### Step4: Solve for $a$
$a=\frac{1650}{3}=550$
### Step5: Solve for $c$
$c=2200-550=1650$
# Answer:
a. System: $\begin{cases} c + a = 2200 \ 6c + 9a = 14850 \end{cases}$
b. 1650 children, 550 adults

---
#### Problem 14
# Explanation:
## Step1: Choose substitution (y isolated)
Substitute $y=x-2$ into $y=-2x+1$
## Step2: Set equal and solve for $x$
$x-2=-2x+1$
$x+2x=1+2$
$3x=3$
$x=1$
## Step3: Substitute $x=1$ into $y=x-2$
$y=1-2=-1$
# Answer:
$x=1, y=-1$ (substitution used because $y$ is already isolated in both equations)

---
#### Problem 15
# Explanation:
## Step1: Choose substitution (isolate $y$)
From $3y+9=3x$, solve for $y$: $y=x-3$
## Step2: Substitute $y=x-3$ into $y=-\frac{1}{3}x+1$
$x-3 = -\frac{1}{3}x +1$
## Step3: Multiply by 3 to eliminate fraction
$3x-9 = -x +3$
## Step4: Solve for $x$
$3x+x=3+9$
$4x=12$
$x=3$
## Step5: Substitute $x=3$ into $y=x-3$
$y=3-3=0$
# Answer:
$x=3, y=0$ (substitution used to isolate $y$ easily)

---
#### Problem 16
# Explanation:
## Step1: Choose substitution (isolate $x$)
From $-x+2y=-4$, solve for $x$: $x=2y+4$
## Step2: Substitute $x=2y+4$ into $4y=x$
$4y=2y+4$
## Step3: Solve for $y$
$4y-2y=4$
$2y=4$
$y=2$
## Step4: Substitute $y=2$ into $x=2y+4$
$x=2(2)+4=8$
# Answer:
$x=8, y=2$ (substitution used to isolate $x$ easily)

---
#### Problem 17
# Explanation:
## Step1: Define variables and equations
Let $a$ = given acute angle, $b$ = other acute angle.
Right triangle acute angles sum to $90^\circ$: $a + b =90$
Given relationship: $a = \frac{1}{4}b$
## Step2: Substitute $a=\frac{1}{4}b$ into sum equation
$\frac{1}{4}b + b =90$
$\frac{5}{4}b=90$
## Step3: Solve for $b$
$b=90	imes\frac{4}{5}=72$
## Step4: Solve for $a$
$a=\frac{1}{4}(72)=18$
# Answer:
The acute angles are $18^\circ$ and $72^\circ$

---
### 5.3 Solving Systems by Elimination
---
#### Problem 18
# Explanation:
## Step1: Add the two equations
$(2x+5y)+(2x-5y)=60+(-20)$
$4x=40$
## Step2: Solve for $x$
$x=\frac{40}{4}=10$
## Step3: Substitute $x=10$ into $2x+5y=60$
$2(10)+5y=60$
$20+5y=60$
$5y=40$
$y=8$
# Answer:
$x=10, y=8$

---
#### Problem 19
# Explanation:
## Step1: Multiply second equation by 3
$3(2x+y)=3(-5) \implies 6x+3y=-15$
## Step2: Add to first equation
$(4x-3y)+(6x+3y)=15+(-15)$
$10x=0$
## Step3: Solve for $x$
$x=0$
## Step4: Substitute $x=0$ into $2x+y=-5$
$0+y=-5$
$y=-5$
# Answer:
$x=0, y=-5$

---
#### Problem 20
# Explanation:
## Step1: Define variables and equations
Let $j$ = jars of jam, $b$ = bread mix packages.
Total items: $j + b =8$
Total cost: $6j +5b=45$
## Step2: Eliminate $b$: multiply first equation by 5
$5j+5b=40$
## Step3: Subtract from cost equation
$(6j+5b)-(5j+5b)=45-40$
$j=5$
## Step4: Substitute $j=5$ into $j+b=8$
$5+b=8$
$b=3$
# Answer:
5 jars of jam, 3 bread mix packages

---
#### Problem 21
# Answer-Explanation Format
# Brief Explanations:
Elimination is easier when:
1. Variables have coefficients that are opposites or easy to scale into opposites, avoiding messy fractions from substitution.
2. Equations are in standard form ($Ax+By=C$) rather than slope-intercept form.
3. Isolating a variable for substitution would result in fractional coefficients.
# Answer:
It is easier to use elimination when the equations have coefficients that can be easily matched to cancel out a variable, or when isolating a variable for substitution would create fractions.

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QUESTION IMAGE

Question

Explanation:

5.2 Solving Systems by Substitution

Problem 10

Step1: Substitute $y=x+9$ into first equation

Step2: Isolate $x$ terms

Step3: Solve for $x$

Step4: Substitute $x=-4$ into $y=x+9$

Step1: Isolate $y$ in first equation

Step2: Substitute $y$ into second equation

Step3: Isolate $x$ terms

Step4: Solve for $x$

Step5: Substitute $x=-8$ into $y=-4-\frac{1}{2}x$

Step1: Isolate $x$ in first equation

Step2: Substitute $x$ into second equation

Step3: Simplify and solve for $y$

Step4: Substitute $y=4$ into $x=28-5y$

Answer:

Problem 11